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Probiem (35marks) 1. Let \( Y \) be the curve \( y=x^{3}-3 x^{2}+4 x-1 \) joining points \( (1 ; 1) \) and \( (2 ; 3) \). Find the value of \( \int_{Y}\left(2 z^{2}-i z\right) d z \)

Ask by Fitzgerald Curry. in Cameroon
Feb 03,2025

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The value of the integral is \( -\frac{73}{3} + \frac{43}{6}i \).

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Mind Expander

To solve the integral \( \int_{Y}(2z^{2}-iz) \, dz \) along the curve \( Y: y = x^{3} - 3x^{2} + 4x - 1 \) from the point \( (1, 1) \) to \( (2, 3) \), we first need to parameterize the curve. We can parameterize the curve using \( z = x + iy \), where \( y = x^3 - 3x^2 + 4x - 1 \). Thus, we have: - \( z = x + i(x^3 - 3x^2 + 4x - 1) \) - \( dz = (1 + i(3x^2 - 6x + 4)) \, dx \) Now we can set up the integral as follows: \[ \int_{1}^{2} \left( 2(x + i(x^{3} - 3x^{2} + 4x - 1))^{2} - i(x + i(x^{3} - 3x^{2} + 4x - 1)) \right) \left( 1 + i(3x^{2} - 6x + 4) \right) \, dx \] This integral is now a bit complex because we have to expand and integrate it term by term, but we can conclude this is a definite integral over the specified bounds. Carry out the detailed calculations and simplifications to obtain your final result. (Note: Because of the components involved, calculations may be tedious but follow through with order and you will arrive at the solution.)

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